$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$
-$$\lim _ { x \rightarrow 0 } \frac { \sin 5 x } { \sin 4 x }$$
A) does not exist
B) $\frac { 4 } { 5 }$
C) $\frac { 5 } { 4 }$
D) 0

Question

Quizplus · Accepted Answer

Using the given limit $\lim _ { x = 0 } \frac { \sin x } { x } = 1$, we can rewrite the given limit as $\lim _ { x \rightarrow 0 } \frac { \sin 5 x } { 5x } \cdot \frac { 4x } { \sin 4 x } \cdot \frac { 5 } { 4 }$. Applying the given limit to both $\frac { \sin 5 x } { 5x }$ and $\frac { 4x } { \sin 4 x }$, both approach 1, leaving us with $\frac { 5 } { 4 }$.

Find the limit using lim⁡x=0sin⁡xx=1. \text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. } Find the limit using limx=0​xsinx​=1.

$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$